In quantum mechanics, it is postulated that to every observable, we have an associated operator. It is further postulated that when we do a measurement on a system, the measured value is one of the eigenvalues of the operator (operator corresponding to the observable), and the wave function (which is initially superposition of many states, say $\psi$) collapses into the eigenfunction ($\phi$) corresponding to the eigenvalue.

Let our operator is $\hat{A}$, and our eigenvalue is $a$

We represent the collapse of wavefunction by this equation

$\hat{A} \phi = a \phi$

But in this equation, I am confused, as the wavefunction is collapsing to eigenfunction upon observation, so the equation should look like this

$\hat{A} \psi = a \phi$

Please help me by telling where I am doing the mistake.

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    $\begingroup$ There is no standard ordering of the postulates of quantum mechanics ;) So, you can drop the descriptions such as "the fifth postulate" because it doesn't mean anything. $\endgroup$ – Dvij D.C. May 15 '20 at 3:24

We represent the collapse of a wavefunction by this equation, $\hat{A} \phi = a \phi$.

This is not true. $\hat{A}\phi=a\phi$ is the eigenvalue equation of the operator $A$. It simply says that the action of the operator $\hat{A}$ on its eigenvector $\phi$ gives $a\phi$ where $a$ is the associated eigenvalue.

The phrase "action of the operator $\hat{A}$ on a state $|\psi\rangle$" can be confusing. It does not mean the physical act of measuring the observable associated with the operator $\hat{A}$. It simply means the mathematical operation of multiplying the given state vector with the said operator.

However, you are correct in expecting that a measurement of an observable associated with the operator $\hat{A}$ on the state $|\psi\rangle$ should be represented somehow! So, how do we represent it? Well, we cannot write down an equation that tells us the end result of the measurement of an operator on a state because of the fact that the result of a measurement in quantum mechanics is fundamentally probabilistic! If we would write something down, it wouldn't remain a surprise as to what would be the outcome of the said measurement, now would it? ;)

However, we can write down some things about the act of measurement. We say that the measurement of an operator $\hat{A}$ on a state $|\psi\rangle$ results in an eigenstate $|\phi_a\rangle$ of the operator $\hat{A}$ with a probability $|\langle \phi_a |\psi\rangle|^2$. How do we write down a process that takes the state $|\psi\rangle$ to the state $|\phi_a\rangle$? Very simple, you project the state $|\psi\rangle$ onto the eigenstate $|\phi_a\rangle$ with the projection operator $\mathbb{P}_a=|\phi_a\rangle\langle \phi_a|$. This is the object you were looking for. You see, $\mathbb{P}_a|\psi\rangle=|\phi_a\rangle\langle \phi_a|\psi\rangle$ which is just $|\phi_a\rangle$ up to normalization.

However, you should notice that $\mathbb{P}_a$ is also not exactly the operator that describes the process of measurement (even aside from the issue of normalization) because it is obviously not certain that the measurement will collapse our initial state to the eigenstate $|\phi_a\rangle$. It can also collapse it to some other eigenstate of $\hat{A}$, say $|\phi_b\rangle$ and if that happens then that process would be described by the action of $\mathbb{P}_b$ on $|\psi\rangle$. So, we can say that the process of the measurement is described by the action of the projection operator $\mathbb{P}_a$ on the state $|\psi\rangle$ with a probability $|\langle\phi_a|\psi\rangle|^2$.

Finally, there is a very nice way to describe the result of a measurement that you haven't yet looked at! Let me explain what I mean. Since the measurement process is fundamentally probabilistic, we obviously cannot actually write down the exact post-measurement state of our system. But, if we have made the measurement but haven't yet looked at the result then we can say that our system is in the eigenstate $|\phi_a\rangle$ with a probability $|\langle \phi_a|\psi\rangle|^2$. And we do have a mathematical object to describe such a system for which we know the probabilities of being in different states. It is called the density matrix. The density matrix of a system which can be in a state $|\lambda_n\rangle$ with a probability $p_n$ is given by $$\hat\rho=\sum_n p_n|\lambda_n\rangle\langle \lambda_n|$$In our case, we know that our system would be in a state $|\phi_a\rangle$ with a probability $|\langle \phi_a|\psi\rangle|^2$. Thus, we represent it with a density matrix \begin{align}\hat{\rho}_\hat{A}&=\sum_a |\langle \phi_a|\psi\rangle|^2 |\phi_a\rangle\langle \phi_a|\\&=\sum_a \langle \psi|\phi_a\rangle\langle\phi_a|\psi\rangle |\phi_a\rangle\langle \phi_a|\\&=\sum_a \langle \psi|\mathbb{P}_a|\psi\rangle \mathbb{P}_a\end{align} where the summation is over the eigenvalues $a$ of the operator $\hat{A}$. The subscript $\hat{A}$ in $\hat{\rho}_\hat{A}$ denotes that the density matrix represents the system after having undergone the measurement of operator $\hat{A}$.

So, to summarize, the act of measurement can be described in two ways.

  • You can say the measurement of $\hat{A}$ on $|\psi\rangle$ gives $\mathbb{P}_a|\psi\rangle$ (up to normalization) with a probability $|\langle\phi_a|\psi\rangle|^2$.
  • If you have made the measurement but haven't looked at the result, you can say that the measurement of $\hat{A}$ on $|\psi\rangle$ has given us a system described by the density matrix $\hat{\rho}_\hat{A}=\sum_a \langle \psi|\mathbb{P}_a|\psi\rangle \mathbb{P}_a$.
  • $\begingroup$ What does "up to normalisation" means? @Dvij D.C. $\endgroup$ – Shine kk May 15 '20 at 4:43
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    $\begingroup$ @Shinekk Notice that $\mathbb{P}_a|\psi\rangle=|\phi_a\rangle\langle\phi_a|\psi\rangle$. So the answer of $\mathbb{P}_a|\psi\rangle$ is not exactly $|\phi_a\rangle$ but $|\phi_a\rangle\langle\phi_a|\psi\rangle$. As you can see $|\phi_a\rangle\langle\phi_a|\psi\rangle$ differs from $|\phi_a\rangle$ only by a multiplicative factor. If you normalize $|\phi_a\rangle\langle\phi_a|\psi\rangle$ (divide the state by its norm to make its norm one) then this multiplicative factor would go away. It's like saying $5\vec{i}$ differs from $\vec{i}$ by a normalization factor. $\endgroup$ – Dvij D.C. May 15 '20 at 11:58
  • $\begingroup$ Okk sir I got it, Earlier I didn't knew that $\langle\phi|\psi\rangle$ is a scalar quantity. $\endgroup$ – Shine kk May 15 '20 at 12:05
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    $\begingroup$ @Shinekk Ah, ok. I guess you now already know about Dirac notations because you figured out that it's a scalar ;) Yes, it's a complex number but just a number, not a vector or a matrix. $\endgroup$ – Dvij D.C. May 15 '20 at 12:09
  • $\begingroup$ Great and pedagocial answer, +1. I'm just a little concerned with the sentence "But, if we have made the measurement but haven't yet looked at the result then we can say that our system is in the eigenstate $|\phi_a \rangle$ with probability $\langle \phi_a | \psi \rangle|^2$." This makes it sound as if it's the "looking" part, and not the measurement part, that alters the state $|\psi \rangle$. $\endgroup$ – Marius Ladegård Meyer May 16 '20 at 6:31

A measurement of the observable $A$ is not represented by applying the operator $A$ to the wavefunction. The transition from $\psi$ to $\phi$ (that is, the observation) cannot be represented by a linear operator, because it's not even a function! As we know, the specific $\phi$ you get is random, so it's not possible to have a mathematical function that tells you what state you get.

What your equation is saying, though, is that if you do the measurement and observe the value $a$, the state $\phi$ after the measurement will satisfy $A\phi = a\phi$. The quantity $A\psi$ is not very relevant here.

  • $\begingroup$ The transition from $\psi$ to $\phi$ can be simply represented by $\mathbb{P}_\phi|\psi\rangle=|\phi\rangle\langle\phi|\psi\rangle$. I obviously agree that it is random as to which specific $\phi$ the state $\psi$ would transition to, but whichever $\phi$ it is, the transition would be given by the action of the projection operator of that $\phi$ on the state $\psi$. $\endgroup$ – Dvij D.C. May 15 '20 at 3:29
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    $\begingroup$ @DvijD.C. Not exactly, because the resulting state is not normalized. You can normalize the result, but then the operator is not linear anymore. $\endgroup$ – Javier May 15 '20 at 3:35
  • $\begingroup$ Ah, I see what you mean. $\endgroup$ – Dvij D.C. May 15 '20 at 3:37

You have discovered the measurement problem in quantum mechanics, and the fact the time evolution is not described by the operator equation shown.

This is what lead Everett to the Many World's interpretation; meanwhile in the more familiar Copenhagen interpretation, wikipedia says,

"The problem [collapse] is deflected by the Copenhagen Interpretation, which postulates that this is a special characteristic of the "measurement" process."

A complete review of all the interpretation is beyond the scope of this answer, but suffice say: your objection is well founded.

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    $\begingroup$ I think this answer misses OP's confusion: the question is about the standard formalism of QM, not about grand conceptual questions such as the measurement problem and quantum interpretations. $\endgroup$ – Javier May 15 '20 at 2:52
  • $\begingroup$ @Javier I don't see any difference in my answer and yours. $\endgroup$ – JEB May 15 '20 at 15:38
  • $\begingroup$ I think the only part of your answer that actually answers OP's question is the the phrase "the time evolution is not described by the operator equation above". My answer explains more clearly what that means, while yours then talks about the measurement problem and interpretations, which I think will only confuse OP more. $\endgroup$ – Javier May 15 '20 at 15:50

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